#include <iostream>
using namespace std;

const int N = 1010;
const int mod = 1e9 + 7;

int n;
int f[N][N];

int main()
{
    cin >> n;
    for (int i = 0; i <= n; ++i)
        f[i][0] = 1;             //可以选某些数，但是容量为0，所以只有一种方式，不选
    for (int i = 1; i <= n; ++i) //可以被选的数
        for (int j = 1; j <= n; ++j)
            for (int k = 0; k * i <= j; ++k)
                f[i][j] = (f[i][j] + f[i - 1][j - k * i]) % mod; //这里不用补
    cout << f[n][n];
    return 0;
}
